Question

Prove that $\cos \left( \dfrac{\pi }{4}+x \right)+\cos \left( \dfrac{\pi }{4}-x \right)=\sqrt{2}\cos x$ .

Answer 1

Hint: Try to simplify the left-hand side of the equation that we need to prove by using the property that $\operatorname{cosA}+cosB=2cos\left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ and put the value of $\cos \left( \dfrac{\pi }{4} \right)$ to reach the required result.

Complete step-by-step answer:

We will now solve the left-hand side of the equation given in the question.
$\cos \left( \dfrac{\pi }{4}+x \right)+\cos \left( \dfrac{\pi }{4}-x \right)$
Now we know $\operatorname{cosA}+cosB=2cos\left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ . If we use the formula in our expression, we get
$2\cos \left( \dfrac{\dfrac{\pi }{4}+x+\dfrac{\pi }{4}-x}{2} \right)\cos \left( \dfrac{\dfrac{\pi }{4}+x-\dfrac{\pi }{4}+x}{2} \right)$
$=2\cos \left( \dfrac{\pi }{4} \right)\cos \left( x \right)$

Now, the value of $\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ . Putting this in our expression, we get
$2\times \dfrac{1}{\sqrt{2}}\times \cos x$
$=\sqrt{2}\cos x$
As we have shown that the left-hand side of the equation given in the question is equal to the right-hand side of the equation in the question. Therefore, we can say that we have proved that $\cos \left( \dfrac{\pi }{4}+x \right)+\cos \left( \dfrac{\pi }{4}-x \right)=\sqrt{2}\cos x$ .

Note: Be careful about the calculation and the signs while opening the brackets. The general mistake that a student can make is 1+x-(x-1)=1+x-x-1. Also, you need to remember the properties related to complementary angles and trigonometric ratios. It is preferred that while dealing with questions as above, you must first try to observe the pattern of the consecutive terms before applying the formulas, as directly applying the formulas may complicate the question.